A rectangular current loop is located near a long straight wire that carries a current of 12A. The current in the loop is 5A. De
1 answer:
Explanation:
The net magnetic force acting on the loop is the superposition of the forces on each side of the loop. The forces on the two sides perpendicular to the long straight wire will cancel (right hand rule). The magnitude of the magnetic force between two wires with currents I_1 = 12 A and I_2 = 5 A of length L = 0.5 m ( assuming not given in question).
so the forces on the near and far wires are
The force F_n is an attractive one, while F_f is a repulsive one (right hand rule). The magnitude of the force is therefore
F = F_n - F_f
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Answer:
the liquid has less height than the mercury
h_{ liquid} =
Explanation:
The pressure as a function of the height is given by
P = ρ g h
where ρ is the density of the liquid, g the acceleration of gravity and h the height reached by the column of the liquid
In that case they say that the pressure is the standard one that is P = 1.01 10⁵ Pa = 760 mmHg
The first way to give the pressure is in SI units and the second way is the height that the mercury column reaches
In the case of building a barometer with a liquid that has a density greater than that of mercury
ρ_liquid > ρ_Hg
the pressure
P =ρ_lquid g h_liquid
if we have the same pressure
ρ_{Hg} g h_{Hg} = ρ_{liquid} g h_{liquid}
h_{ liquid} =
therefore the liquid has less height than the mercury
(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to
where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and is the variation of height. Substituting the numbers into the formula, we find
(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:
However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.
(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):